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Math Help - A shortcut to the Maclaurin series expansion?

  1. #1
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    A shortcut to the Maclaurin series expansion?

    (Not sure if this should go into the calculus section; feel free to move it.)

    Question: Find the Maclaurin series for (p+qx^2)^r up to and including the term in x^4 where p, q, r \in \mathbb{R}

    The key gives the following answer:
    p^r(1+\frac{p}{q}x^2)=p^r(1+r\frac{q}{p}x^2+\frac{  r(r-1)}{2}\frac{q^2}{p^2}x^4)

    I understand how they can move out p^r and expand the parentheses binomially to get the right hand side of the equation, but there are two things that puzzle me:

    1. The question asks for me to find a Maclaurin series for the expression. To me, that means finding the first derivatives and of the expression, dividing them by a factorial and multiplying by an x^n term. This is not done here. Why does the answer given correspond to a Maclaurin series without performing these steps?
    2. In which cases is it valid to do the above trick? When should one think of using it?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by mtpastille View Post
    Question: Find the Maclaurin series for (p+qx^2)^r up to and including the term in x^4 where p, q, r \in \mathbb{R}

    The key gives the following answer:
    p^r(1+\frac{p}{q}x^2)=p^r(1+r\frac{q}{p}x^2+\frac{  r(r-1)}{2}\frac{q^2}{p^2}x^4)

    I understand how they can move out p^r and expand the parentheses binomially to get the right hand side of the equation, but there are two things that puzzle me:

    1. The question asks for me to find a Maclaurin series for the expression. To me, that means finding the first derivatives and of the expression, dividing them by a factorial and multiplying by an x^n term. This is not done here. Why does the answer given correspond to a Maclaurin series without performing these steps?
    Bad terminology. I would call it a binomial series, not a Maclaurin series.

    Quote Originally Posted by mtpastille View Post
    2. In which cases is it valid to do the above trick? When should one think of using it?
    Except in the elementary case where r is a positive integer, this binomial expansion will only converge when \bigl|\tfrac qpx^2\bigr|<1. That is probably the only situation when you would want to use this result.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mtpastille View Post
    2. In which cases is it valid to do the above trick? When should one think of using it?

    Always

    (1+u)^r=\displaystyle\sum_{n=0}^{+\infty}\binom{r}  {n}u^n \quad (|u|<1,\;r\in \mathbb{R})


    Edited: Sorry, I didn't see Opalg's post.
    Last edited by FernandoRevilla; May 6th 2011 at 12:05 PM.
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